What methods, not including the one of characteristic curves and surfaces, can I use to solve first-order quasilinear equations like this one?

$${u}_{x}+{u}_{y}={u}^{2}$$

$${u}_{x}+{u}_{y}={u}^{2}$$

eukrasicx
2022-09-29
Answered

What methods, not including the one of characteristic curves and surfaces, can I use to solve first-order quasilinear equations like this one?

$${u}_{x}+{u}_{y}={u}^{2}$$

$${u}_{x}+{u}_{y}={u}^{2}$$

You can still ask an expert for help

Matteo Estes

Answered 2022-09-30
Author has **9** answers

$\frac{dx}{dt}}=1$ , letting x(0)=0 , we have x=t

$\frac{dy}{dt}}=1$ , letting $y(0)={y}_{0}$ , we have $y=t+{y}_{0}=x+{y}_{0}$

$\frac{du}{dt}}={u}^{2$ , we have $u(x,y)={\displaystyle \frac{1}{f({y}_{0})-t}}={\displaystyle \frac{1}{f(y-x)-x}}$

$\frac{dy}{dt}}=1$ , letting $y(0)={y}_{0}$ , we have $y=t+{y}_{0}=x+{y}_{0}$

$\frac{du}{dt}}={u}^{2$ , we have $u(x,y)={\displaystyle \frac{1}{f({y}_{0})-t}}={\displaystyle \frac{1}{f(y-x)-x}}$

asked 2022-09-22

What is a solution to the differential equation $\frac{dy}{dx}=1+2xy$?

asked 2022-06-18

I need to convert the following differential equation to standard form.

${T}_{n}=2{T}_{n-1}+1$

(not quite sure how to really format it properly)

I was thinking it is

${T}_{n}-2{T}_{n-1}-1$

If I'm incorrect (or correct) can I please have somebody just explain briefly about the decreasing order. I'm just confused about the whole number, in this case the number 1, I'm not quite sure if it's bigger or smaller than the others.

${T}_{n}=2{T}_{n-1}+1$

(not quite sure how to really format it properly)

I was thinking it is

${T}_{n}-2{T}_{n-1}-1$

If I'm incorrect (or correct) can I please have somebody just explain briefly about the decreasing order. I'm just confused about the whole number, in this case the number 1, I'm not quite sure if it's bigger or smaller than the others.

asked 2022-06-01

How can I start to solve this differential equation?

${y}^{\prime}=\frac{y}{2y\mathrm{ln}(y)+y-x}$

${y}^{\prime}=\frac{y}{2y\mathrm{ln}(y)+y-x}$

asked 2022-09-04

What is a general solution to the differential equation $y\prime =5{x}^{\frac{2}{3}}{y}^{4}$?

asked 2022-05-15

I am given this:

$(2x+1)\frac{dy}{dx}+y=0$

I tried this:

$\frac{1}{(2x+1)}dx=\frac{-1}{y}dy$

Then integrated the above sum and got this:

$\frac{ln(2x+1)}{2}=-ln(y)$

The answer is: ${y}^{2}(2x+1)=C$.

I tried solving it by placing the like terms together and integrating them. However, my answer is wrong from the answer given. Could you point out my mistake? Or am i evaluating the entire thing incorrectly?

All suggestions and help are appreciated!

$(2x+1)\frac{dy}{dx}+y=0$

I tried this:

$\frac{1}{(2x+1)}dx=\frac{-1}{y}dy$

Then integrated the above sum and got this:

$\frac{ln(2x+1)}{2}=-ln(y)$

The answer is: ${y}^{2}(2x+1)=C$.

I tried solving it by placing the like terms together and integrating them. However, my answer is wrong from the answer given. Could you point out my mistake? Or am i evaluating the entire thing incorrectly?

All suggestions and help are appreciated!

asked 2022-02-15

How is ${y}^{\prime}\left(x\right)=f(x,y\left(x\right))$ is the most general first order ordinary differential equation ? Isn't $f(x,y\left(x\right),{y}^{\prime}\left(x\right))=0$ the most general first order ODE ? I mean isn't a restriction to single out ${y}^{\prime}\left(x\right)$ from $f$ ? Thank you for your help!

asked 2022-07-07

Problem:

Solve the following differential equation:

$\begin{array}{rcl}6{x}^{2}y\phantom{\rule{thinmathspace}{0ex}}dx-({x}^{3}+1)\phantom{\rule{thinmathspace}{0ex}}dy& =& 0\end{array}$

Answer:

This is a separable differential equation.

$\begin{array}{rcl}\frac{6{x}^{2}}{{x}^{3}+1}\phantom{\rule{thinmathspace}{0ex}}dx-\frac{dy}{y}& =& 0\\ \int \frac{6{x}^{2}}{{x}^{3}+1}\phantom{\rule{thinmathspace}{0ex}}dx-\int \frac{dy}{y}& =& {c}_{1}\\ 2\mathrm{ln}|{x}^{3}+1|-\mathrm{ln}|y|& =& {c}_{1}\\ \mathrm{ln}({x}^{3}+1{)}^{2}-\mathrm{ln}|y|& =& {c}_{1}\\ \mathrm{ln}{\textstyle (}\frac{({x}^{3}+1{)}^{2}}{|y|}{\textstyle )}& =& {c}_{1}\\ ({x}^{3}+1{)}^{2}& =& c|y|\end{array}$

However, the book gets:

$\begin{array}{rcl}({x}^{3}+1{)}^{2}& =& |cy|\end{array}$

Is my answer different from the book's answer? I believe it is.

Solve the following differential equation:

$\begin{array}{rcl}6{x}^{2}y\phantom{\rule{thinmathspace}{0ex}}dx-({x}^{3}+1)\phantom{\rule{thinmathspace}{0ex}}dy& =& 0\end{array}$

Answer:

This is a separable differential equation.

$\begin{array}{rcl}\frac{6{x}^{2}}{{x}^{3}+1}\phantom{\rule{thinmathspace}{0ex}}dx-\frac{dy}{y}& =& 0\\ \int \frac{6{x}^{2}}{{x}^{3}+1}\phantom{\rule{thinmathspace}{0ex}}dx-\int \frac{dy}{y}& =& {c}_{1}\\ 2\mathrm{ln}|{x}^{3}+1|-\mathrm{ln}|y|& =& {c}_{1}\\ \mathrm{ln}({x}^{3}+1{)}^{2}-\mathrm{ln}|y|& =& {c}_{1}\\ \mathrm{ln}{\textstyle (}\frac{({x}^{3}+1{)}^{2}}{|y|}{\textstyle )}& =& {c}_{1}\\ ({x}^{3}+1{)}^{2}& =& c|y|\end{array}$

However, the book gets:

$\begin{array}{rcl}({x}^{3}+1{)}^{2}& =& |cy|\end{array}$

Is my answer different from the book's answer? I believe it is.